# LeetCode 1441. Build an Array With Stack Operations

## Description

https://leetcode.com/problems/build-an-array-with-stack-operations/

Given an array `target` and an integer `n`. In each iteration, you will read a number from  `list = {1,2,3..., n}`.

Build the `target` array using the following operations:

• Push: Read a new element from the beginning `list`, and push it in the array.
• Pop: delete the last element of the array.
• If the target array is already built, stop reading more elements.

Return the operations to build the target array. You are guaranteed that the answer is unique.

Example 1:

```Input: target = [1,3], n = 3
Output: ["Push","Push","Pop","Push"]
Explanation:
Read number 1 and automatically push in the array -> 
Read number 2 and automatically push in the array then Pop it -> 
Read number 3 and automatically push in the array -> [1,3]
```

Example 2:

```Input: target = [1,2,3], n = 3
Output: ["Push","Push","Push"]
```

Example 3:

```Input: target = [1,2], n = 4
Output: ["Push","Push"]
Explanation: You only need to read the first 2 numbers and stop.
```

Example 4:

```Input: target = [2,3,4], n = 4
Output: ["Push","Pop","Push","Push","Push"]
```

Constraints:

• `1 <= target.length <= 100`
• `1 <= target[i] <= n`
• `1 <= n <= 100`
• `target` is strictly increasing.

## Explanation

Track the numbers added and compare with the target array to decide whether “push” or “pop” operation to use.

## Python Solution

``````class Solution:
def buildArray(self, target: List[int], n: int) -> List[str]:
results = []
numbers = []

for i in range(1, n + 1):
if numbers == target:
return results

numbers.append(i)

if numbers[len(numbers) - 1] == target[len(numbers) - 1]:
results.append("Push")
else:
results.append("Push")
results.append("Pop")
numbers.pop()

return results
``````
• Time Complexity: O(N).
• Space Complexity: O(N).